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3 years ago

What is true of a reaction that has reached equilibrium?

2 answers:

8 0

I believe the correct answer from the choices listed above is option A. A reaction that has reached equilibrium would mean that the reaction rates of the forward and reverse reactions are equal. Hope this answers the question. Have a nice day.

BabaBlast [244]3 years ago

4 0

The correct option is A.

A chemical reaction is said to have reached an equilibrium stage if the rate of reaction of the forward reaction is equal to the rate of reaction of the reverse reaction. Two way arrows are usually used to depict equilibrium reactions. These arrows indicate that the chemical reaction can move both ways. At the equilibrium point the concentrations of both the reactants and the products are equal.

You might be interested in

Write balanced chemical equation for the reactions used to prepare each of the following compounds from the given starting mater

creativ13 [48]

Explanation:

A balanced chemical equation is defined as the one that contains same number of atoms on both reactant and product side.

Whereas a chemical reaction that brings change in oxidation state of the reacting species is known as a redox reaction.

When there occurs decrease in oxidation number of an element then it shows the element has been reduced. And, if there occurs an increase in oxidation state of an element then it means the element has been oxidized.

(a) The reaction between nitrogen and hydrogen to form ammonia will be as follows.

$N_{2} + 3H_{2} \rightarrow 2NH_{3}$

So here, oxidation state of nitrogen is changing from 0 to -3. Whereas oxidation state of hydrogen is changing from 0 to +3.

Reaction between ammonia and nitric acid will be as follows.

$NH_{3} + HNO_{3} \rightarrow NH_{4}NO_{3}$

Since, it is a combination reaction. Therefore, no change in oxidation state of reacting species is taking place.

(b) Reaction for conversion of liquid bromine to hydrogen bromide is as follows.

$H_{2}(g) + Br_{2}(l) \rightarrow 2HBr(g)$

Here, oxidation state of hydrogen is changing from 0 to +1. Therefore, oxidation of hydrogen is taking place. On the other hand, oxidation state of bromine is changing from zero to -1. Therefore, reduction of bromine is taking place.

(c) Reaction between Zn and S will be as follows.

$Zn(s) + S(s) \rightarrow ZnS(s)$

Here, oxidation state of zinc is changing from 0 to +2. Hence, oxidation of zinc is taking place.

Oxidation state of sulfur is changing from 0 to -2. Therefore, reduction of sulfur is taking place.

7 0

3 years ago

Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first i

Law Incorporation [45]

Answer:

a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹

c. 2.3 × 10⁻⁴ mol·L⁻¹; 5.5 × 10⁻⁸ mol·L⁻¹

Explanation:

a. Silver iodate

Let s = the molar solubility.

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹: s s

$K_{sp} =\text{[Ag$^{+}$][IO$_{3}$$^{-}$]} = s\times s = s^{2} = 3.0\times 10^{-8}\\s = \sqrt{3.0\times 10^{-8}} \text{ mol/L} = 1.7 \times 10^{-4} \text{ mol/L}$

b. Barium sulfate

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹: 0.02 0

C/mol·L⁻¹: +s +s

E/mol·L⁻¹: 0.02 + s s

$K_{sp} =\text{[Ba$^{2+}$][SO$_{4}$$^{2-}$]} = (0.02 + s) \times s \approx 0.02s = 1.1\times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{0.02} \text{ mol/L} = 5.5 \times 10^{-9} \text{ mol/L}$

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is

$\mu = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\\mu = \dfrac{1}{2} (\text{[Ba$^{2+}$]}\cdot (2+)^{2} + \text{[NO$_{3}$$^{-}$]}\times(-1)^{2}) = \dfrac{1}{2} (\text{0.02}\times 4 + \text{0.04}\times1)= \dfrac{1}{2} (0.08 + 0.04)\\\\= \dfrac{1}{2} \times0.12 = 0.06$

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(1)^{2}\sqrt{0.06} = -0.51\times 0.24 = -0.12\\\gamma = 10^{-0.12} = 0.75$

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

$K_{sp} =\text{[Ag$^{+}$]$\gamma_{Ag^{+}}$[IO$_{3}$$^{-}$]$\gamma_{IO_{3}^{-}}$} = s\times0.75\times s \times 0.75 =0.56s^{2}= 3.0 \times 10^{-8}\\s^{2} = \dfrac{3.0 \times 10^{-8}}{0.56} = 5.3 \times 10^{-8}\\\\s =2.3 \times 10^{-4}\text{ mol/L}$

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(2)^{2}\sqrt{0.06} = -0.51\times16\times 0.24 = -0.50\\\gamma = 10^{-0.50} = 0.32$

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

$K_{sp} =\text{[Ba$^{2+}$]$\gamma_{ Ba^{2+}}$[SO$_{4}$$^{2-}$]$\gamma_{ SO_{4}^{2-}}$} = (0.02 + s) \times 0.32\times s\times 0.32 \approx 0.02\times0.10s\\2.0\times 10^{-3}s = 1.1 \times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{2.0 \times 10^{-3}} \text{ mol/L} = 5.5 \times 10^{-8} \text{ mol/L}$

7 0

3 years ago

These models show the electron structures of two different nonmetal elements.

Soloha48 [4]

Answer: Option (D) is the correct answer.

Explanation:

Valence shell is the shell present on the outermost core of an atom and electrons present in the valence shell are known as valence electrons.

If an atom has completely filled valence shell then it means the atom is not reactive in nature because it is already stable.

But when an atom has less than eight electrons in its valence shell then it means to attain stability the atom will readily attract electrons towards itself.

As the given element 1 has 8 electrons in its valence shell. Hence, it is not reactive in nature but element 2 has 6 valence electrons. So, in order to attain stability element 2 will readily attract 2 electrons from a donor atom.

Thus, we can conclude that element 2 is more reactive because it does not have a full valence shell, so it will attract electrons.

4 0

3 years ago

The time it takes to run a kilometer depends on the amount of exercise a person gets.

Debora [2.8K]

<span>The amount of exercise that a person gets would be the independent variable, as it would impact the amount of time it would take to run the kilometer, with time being the dependent variable.</span>

3 0

3 years ago

The pressure is changed from 500 kPa to 250 kPa. What would you expect the new volume to be if the initial volume is 200 mL?

Romashka [77]

Answer:

<h3>The answer is 400 mL</h3>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the new volume

$V_2 = \frac{P_1V_1}{P_2} \\$

So we have

$V_2 = \frac{500000 \times 200}{250000} = \frac{100000000}{250000} = \frac{10000}{25} \\$

We have the final answer as

Hope this helps you

4 0

3 years ago